Question

proving the slope criteria
the slope of \\( \overleftrightarrow { pq } \\) is \\( ( y - s ) \\) divided by
the slope of \\( \overleftrightarrow { p ^ { prime } q ^ { prime } } \\) isdivided by \\( ( x + a ) - ( x - a ) \\)
both lines have a slope that isdivided by
therefore, the lines are

Explanation:

Step1: Calculate the slope of \(\overrightarrow{PQ}\)

The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For points \(P(x,a)\) and \(Q(x - a,x + b)\), the slope of \(\overrightarrow{PQ}\) is \((x + b-a)\) divided by \((x - a)-x=-a\).

Step2: Calculate the slope of \(\overrightarrow{P'Q'}\)

For points \(P'(x - a,a - b)\) and \(Q'(x,x)\), the slope of \(\overrightarrow{P'Q'}\) is \((x-(a - b))\) divided by \((x-(x - a))\). Simplify \(x-(a - b)=x - a + b\) and \((x-(x - a))=a\).

Step3: Simplify the slopes

The slope of \(\overrightarrow{PQ}=\frac{(x + b-a)}{-a}=\frac{-x - b + a}{a}\), and the slope of \(\overrightarrow{P'Q'}=\frac{x - a + b}{a}\). Both lines have a slope that is \((b - a + x)\) divided by \(a\) (after re - arranging terms for \(\overrightarrow{PQ}\) slope: \(\frac{x + b - a}{-a}=-\frac{x + b - a}{a}\), and for \(\overrightarrow{P'Q'}\) slope \(\frac{x-(a - b)}{a}=\frac{x - a + b}{a}\)). Since the slopes of the two lines are equal (after proper algebraic manipulation and considering the general form of slope formula for the two line - segments formed by the given points), the lines are parallel.

Answer:

The slope of \(\overrightarrow{PQ}\) is \((x + b - a)\) divided by \(-a\). The slope of \(\overrightarrow{P'Q'}\) is \((x-(a - b))\) divided by \((x-(x - a))\). Both lines have a slope that is \((b - a + x)\) divided by \(a\) (after algebraic manipulation). Therefore, the lines are parallel.